Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is $$ f=P_0+\cfrac{1}{P_1+\cfrac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.

Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is $$ f=P_0+\cfrac{1}{P_1+\cfrac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.

Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is $$ f=P_0+\frac{1}{P_1+\frac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.

Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is $$ f=P_0+\cfrac{1}{P_1+\cfrac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.